A circle of radius \(R\) is centered at the origin. The shaded top portion is bounded above by the circle and below by the horizontal chord through the point where the radius makes \(60^\circ\) with the \(\boldsymbol{y}\)-axis (see figure). The solid formed by a complete rotation of this shaded part about the \(y\)-axis has volume \(k\pi R^{3}\). Find \(k\).
Show Hint
For a circular segment revolved about the circle's symmetry axis, think "spherical cap." Memorize \(V=\dfrac{\pi h^{2}}{3}(3R-h)\); it saves setting up integrals.
Step 1: Identify the spherical cap height. The chord passes through the point on the circle at a central angle \(60^\circ\) from the \(y\)-axis, i.e. the \(y\)-coordinate is \[ y_0 = R\cos 60^\circ = \frac{R}{2}. \] The shaded part above this chord becomes, upon revolution about the \(y\)-axis, a spherical cap of height \[ h = R - y_0 = R - \frac{R}{2} = \frac{R}{2}. \]
Step 2: Use the spherical-cap volume formula. For a sphere of radius \(R\), a cap of height \(h\) has volume \[ V = \frac{\pi h^{2}}{3}\,(3R - h). \] With \(h=\dfrac{R}{2}\), \[ V = \frac{\pi}{3}\left(\frac{R^{2}}{4}\right)\left(3R - \frac{R}{2}\right) = \frac{\pi R^{3}}{12}\left(\frac{6-1}{2}\right) = \frac{5\pi R^{3}}{24}. \] Step 3: Read off \(k\). Comparing \(V=k\pi R^{3}\) gives \(k=\dfrac{5}{24}\). \[ \boxed{\dfrac{5}{24}} \]
